I designed a PCB where I have a local oscillator Input to an IC. The LO signal (2.48 GHz) has a power of -20 dBm. I put a cut SMA cable on the ground plane and measured a signal level of -60dBm. Is this normal?
\$\begingroup\$
\$\endgroup\$
Add a comment
|
1 Answer
\$\begingroup\$
\$\endgroup\$
4
At 2.5 GHz, the wavelength in free space is only about 120 mm or 5 inches. That means if your distances exceed about 12 mm or a half inch, you don't have a lumped system anymore.
It's not clear what exactly you did, but it seems that you are measuring some signal at -40 dB where you expected none at all. First, none is never realistic. Second, this not being a lumped system, exactly how you measured it is significant.
The fact that a 2.5 GHz signal shows up somewhere else nearby attenuated by 40 dB doesn't by itself seem surprising.
You say you measured this "on the ground plane". This brings up the question of how exactly you measured this, which you haven't stated. You also seem to be under the false assumption that every place on the ground plane should be at the same voltage all the time. That's the desire, and is often a useful approximation, but at 2.5 GHz you can't just wave your hand at such things. The exact geometry matters, as does the route of the return current to this oscillator.
answered Jan 29, 2018 at 13:30
-
\$\begingroup\$ First thank you for your explanation. I measured this with a spectrum analyzer by just cutting a SMA cable and only putting the signal trace to a via which goes to ground. It could be that the signal trace pin of the cable is capturing the EM field in the air. \$\endgroup\$– OcKCommented Jan 29, 2018 at 13:50
-
2\$\begingroup\$ @OcK, what did you connect the outer conductor of the cable to? \$\endgroup\$ Commented Jan 29, 2018 at 15:37
-
1\$\begingroup\$ @ThePhoton I Connected the outer conductor to ground (SMA socket) on the PCB. But the difference to measuring without connecting it to ground is only 3 dB ...:-/ \$\endgroup\$– OcKCommented Jan 29, 2018 at 16:05
-
\$\begingroup\$ So outer conductor is connected to ground at one point on the PCB, and center conductor is connected to ground at another point? You're measuring either IR drop due to ground currents between those two points (less likely), or magnetic fields in the loop between the two connections to ground (more likely). \$\endgroup\$ Commented Jan 29, 2018 at 16:50